Set $S= \mathbb{Q} \times \mathbb{Q}^{*}$ with the binary operation $(i,j)\star (v,w)=(iw+v, jw)$

Question

Consider the set $S= \mathbb{Q} \times \mathbb{Q}^{*}$ with the binary operation $(i,j)\star (v,w)=(iw+v, jw)$.

(a) Show that the binary operation is closed. I said the operation is closed under addition because when you add two rational numbers your sum is a rational number. I'm not sure if there is more to show for this

(b) she that $S$ with the binary operation defined above is a group. I know in order to be a group you need identity element. associative property, and inverse so I have: having the identity element is a structural property of a binary operation therefore $S$ is non empty. $S$ is also associative because $(i,j)\star (v,w)\star (x,y)=(ivx+ix+vx+x, jwy)$ and $(ivx+ix+vx+x, jwy)=(i,j)\star (v,w)\star (x,y)$. and I am stuck on how to show the inverse.

(c) Is $S'= \mathbb{Z} \times \mathbb{Z}^{*}$ with the same binary operation $\star$ a group? I said no because it isn't associative but I'm not sure if thats correct either

Answer 1

(a) Yes, what you wrote is relevant, although sketchy. You also didn't address why you think $jw\in \mathbb Q^\ast$, which would be the last piece. You should really take some time to write out more clearly what you mean.

(b) I know in order to be a group you need identity element. associative property, and inverse That is important, but after that you begged the question that it has an identity and that it is associative and didn't write anything meaningful. You should write out what it means for $(i,j)\star (v,w)=(iw+v, jw)=(v,w)$ to uncover the identity, and then you should write out what it means for $(i,j)\star (v,w)=(iw+v, jw)=Id$ once you know what the identity $Id$ is. In this way you will prove there is an identity, and you can proceed to prove what the inverses are, and determine whether or not it is associative.

(c) I said no because it isn't associative but I'm not sure if thats correct either I'm not sure what possesses you to spout answers without having some justification, however incomplete. You will probably find it's closed under multiplication using the same logic as before, but I think you will find it lacks inverses. You should solve the first two questions thoroughly and then you should find it easy to do (c) with this hint.

Answer 2

You need to justify your statements.

The set $S$ is closed under the operation, because for $(i,j),(v,w)\in S$, the pair $(v+iw,jw)$ belongs to $S$: indeed, $v+iw\in\mathbb{Q}$ and $jw\in\mathbb{Q}^*$ because both $j$ and $w$ are nonzero, so their product is nonzero as well.

A set with an operation can be nonempty, but have no identity element. Consider the set $X$ of even integers, equipped with multiplication; it is closed, but has no identity.

The set $S$ has an identity with respect to $\star$: verify that $(0,1)$ satisfies the requirement.

A way to find it is to solve the equations $v+iw=v$ and $jw=w$. These must hold for every $(v,w)\in S$, forcing $i=0$ and $j=1$. Next you can verify that $(0,1)$ (the only possible identity) is indeed an identity.

Associativity does hold: your task is to expand with the definitions and check that, for all $(i,j),(v,w),(x,y)\in S$, $$ (i,j)\star\bigl((v,w)\star(x,y)\bigr)= \bigl((i,j)\star(v,w)\bigr)\star(x,y) $$

In order to find the inverse of $(v,w)$, you look for $(i,j)$ such that $(i,j)\star(v,w)=(0,1)$ and verify that for the element you find also $(v,w)\star(i,j)=(0,1)$ holds.

What fails in the case of $S'=\mathbb{Z}\times\mathbb{Z}^*$? Repeat the steps above and the answer will be clear.

Answer 3

I said the operation is closed under addition because when you add two rational numbers your sum is a rational number. I'm not sure if there is more to show for this

⋆ is built up out of multiplication and addition, so you need to show that both constituent operations are close.

having the identity element is a structural property of a binary operation therefore S is non empty.

Neither the premise, nor the relevance of the conclusion, nor the logic that ties the premise to conclusion, is clear to me for that attempted syllogism.

S is also associative because (i,j)⋆(v,w)⋆(x,y)=(ivx+ix+vx+x,jwy) and (ivx+ix+vx+x,jwy)=(i,j)⋆(v,w)⋆(x,y).

This is such a non sequitur that I have to wonder whether you understand what "associative" means. Can you in your question quote a definition of "associative" and explain in your own words what it means?

I said no because it isn't associative but I'm not sure if thats correct either

An operation can't go from associative to non-associative by removing elements. If there were integers for which it is non-associative, then those integers would be in $\mathbb{Q} \times \mathbb{Q}^{*}$, so associativity would fail in $\mathbb{Q} \times \mathbb{Q}^{*}$. You want to look for a property that could fail by removing elements (hint: if a property is of the form "there is an element with this property", then removing elements could leave you without any elements with that property).